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V is a finite dimensional vector space. How to prove that any linear map on subspace of V can be extended to linear map on V.

I attempted it by taking the basis of the subspace W with Dim(m) and extending it to basis of V with Dim(n). Since mapping of all vectors of basis of W exists and taking the remaining m-n vectors as zero - we will get a linear map to an element that is in V. But is it sufficient to prove that all vectors in V will also be having a linear map?

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If $\{w_1,\ldots,w_m\}$ is a basis of $W$, all you need to be able to do is show that there exist vectors $v_{m+1},\ldots,v_n \in V$ with the property that $\{w_1,\ldots,w_m,v_{m+1},\ldots,v_n\}$ is a basis of $V$. This is a pretty standard result in linear algebra.

The extension you propose $$\tilde L (v) = \tilde L (\alpha_1 w_1 + \cdots + \alpha_m w_m + \alpha_{m+1}v_{m+1} + \cdots \alpha_n v_n) = L(\alpha_1 w_1 + \cdots + \alpha_m w_m)$$ is linear and uniquely defined for all $v \in V$.

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@ Umberto Please correct me if I'm wrong but a linear map on V should be defined for all elements of V. But does the extension in the answer not limit to vectors in V with $ v_{m+1}=.....=v_n= 0 $ thereby excluding vectors in V where these are not zero –  Hashtag Sep 12 '13 at 18:31
    
Since $\{w_1,\ldots,w_m,v_{m+1},\ldots,v_n\}$ is a basis of $V$, every vector $v \in V$ has a unique representation $v = \alpha_1 w_1 + \cdots + \alpha_m w_m + \alpha_{m+1}v_{m+1} + \cdots \alpha_n v_n$. –  Umberto P. Sep 13 '13 at 13:06
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